3
$\begingroup$

I have trouble understanding the proof of Proposition 2.18 in Karatzas/Shreve: Brownian Motion and Stochastic Calculus, which states that a $\mathcal{F_t}$-progressively measurable process $X_t$, stopped at a $\mathcal{F_t}$-stopping time $\tau$, remains progressive, i. e. $X_{t\wedge\tau}$ is $\mathcal{F_t}$-progressive.

The main idea is to decompose the map $[0,T]\times\Omega\to\mathbb{R}^d, (t,\omega)\mapsto X_{t\wedge\tau(\omega)}(\omega)$ into $M: (t,\omega)\mapsto (t\wedge\tau(\omega),\omega)$ and $(t,\omega)\mapsto X_t(\omega)$ (for all $T>0$). Then it suffices to show that $M$ is $(\mathcal{B}[0,T]\otimes \mathcal{F}_T)$-measurable, since the other map is measurable because of the assumption of progressive measurability of $X$.

The main idea is clear to me. But I don’t see how $M$ is measurable. It seems intuitively clear to me (thinking of the filtration as information), but I don’t manage to prove it formally.

$\endgroup$
6
$\begingroup$

Recall that the product sigma-algebra $\mathcal{B}[0,T] \otimes \mathcal{F}_T$ is generated by

$$\{[0,t] \times B; t \leq T, B \in \mathcal{F}_T\}.$$

Consequently, it suffices to prove that $M^{-1}([0,t] \times B) \in \mathcal{B}[0,T] \otimes \mathcal{F}_T$ for $t \leq T$, $B \in \mathcal{F}_T$. This follows from

$$\begin{align*} M^{-1}([0,t] \times B) &= \{(s,\omega) \in [0,T] \times \Omega; s \wedge \tau(\omega) \leq t, \omega \in B\} \\ &= \{(s,\omega); s \wedge \tau(\omega) \leq t, \omega \in B \cap [\tau \leq t]\} \cup \{(s,\omega); s \wedge \tau(\omega) \leq t, \omega \in B \cap [\tau>t]\} \\ &= ([0,T] \times (B \cap[\tau \leq t])) \cup ([0,t] \times (B \cap [\tau>t])).\end{align*}$$

$\endgroup$
  • $\begingroup$ Thank you for your comment. I didn’t know it suffices to check the measurability on the generator set. I used a similar construction as yours for arbitrary sets which involved an uncountable union, which was very hard to approximate. The “generator lemma” is very nice and was straight-forward to prove. $\endgroup$ – Fye Oct 26 '14 at 11:33
  • $\begingroup$ @Fye Glad I could help you. Actually, this "generator lemma" is used quite often in probability theory. $\endgroup$ – saz Oct 26 '14 at 21:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.